Question

A solid metal sphere of radius 4.00 m carries a total charge of -5.20 μC . a What is the magnitude of the electric field at a distance from the sphere's center of 0.300mb What is the magnitude of the electric field at a distance from the sphere's center of 3.85mc What is the magnitude of the electric field at a distance from the sphere's center of 4.10md What is the magnitude of the electric field at a distance from the sphere's center of 6.00m

Answer 1

a. In a metal sphere, all the charge resides on the surface of the sphere, such that the electric field anywhere in the bulk of the metal sphere is 0, i.e. electric field inside the radius of the metal sphere is 0.

Hence,

$\vec{E}(r=0.300m)=\vec{0}$

b. By same argument,

$\vec{E}(r=3.85m)=\vec{0}$

c. But outside the metal sphere, the electric field will be,

$\vec{E}(r)=\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\hat{r}$

here Q = -5.20×10^(-6) C, r = 4.10m,

$E(r=4.10m)=9\times10^9\times\frac{(5.20\cdot10^{-6})}{(4.10)^2}=2784.06\text{ N/C}$

d. Again outside the sphere,

$E(r=6.00m)=9\times10^9\times\frac{(5.20\times10^{-6})}{(6.00)^2}=1300\text{ N/C}$

Result: a. 0, b. 0, c. 2784.06 N/C, d. 1300 N/C.

In case of solid nonconductor uniformly charged, the electric field won't be 0 inside,

$E(r)=\frac{\rho r}{3\epsilon_0}$

where rho is the charge density.

a.

$E(r=0.300m)=\frac{1}{4\pi\epsilon_0}\frac{Qr}{R^3}=9\times10^9\times\frac{(5.2\times10^{-6})(0.300)}{(4.00)^3}=219.375\text{ N/C}$

b.

$E(r=3.85m)=9\times10^9\times\frac{(5.2\times10^{-6})(3.85)}{(4.00)^3}=2815.3125\text{ N/C}$